To complete the missing parts of the table for the given function $y = \left(\frac{1}{2}\right)^x$, let's calculate $y$ for the missing values of $x$:

Function:

$y = \left(\frac{1}{2}\right)^x$

Step-by-Step Calculations:

1. At $x = -2$: $y = \left(\frac{1}{2}\right)^{-2} = \left(\frac{2}{1}\right)^2 = 2^2 = 4$ So, $y = 4$.

2. At $x = 0$: $y = \left(\frac{1}{2}\right)^0 = 1$ So, $y = 1$.

3. At $x = 3$: $y = \left(\frac{1}{2}\right)^3 = \frac{1}{2^3} = \frac{1}{8}$ So, $y = \frac{1}{8}$.

Completed Table:

Final Answers:

• At $x = -2$, $y = 4$
• At $x = 0$, $y = 1$
• At $x = 3$, $y = \frac{1}{8}$

Let me know if you'd like more explanations!

Related Questions:

1. What happens to $y$ as $x \to \infty$ for this function?
2. How does the graph of $y = \left(\frac{1}{2}\right)^x$ look?
3. What is the behavior of the function for negative values of $x$?
4. How does changing the base to $\frac{1}{3}$ affect the table?
5. What is the relationship between exponential growth and decay?

Tip:

For any $a^x$ where $0 < a < 1$, the function represents exponential decay because the values of $y$ get smaller as $x$ increases.